Function/Set Tough problem

[bala977]

I took a GMATPrep yesterday. Encountered many tough questions - one of them was

" if N is a number such that h(N) equals the product of all even integers from 2 to N , then what could be the least prime factor of h(100)+1? "

I think the answer choices were

1) Between 0 to 10
2) Between 10 to 20
3) Between 20 to 30
4) Between 30 to 40
5) Greater than 40

SPOILER: The answer was greater than 40

Can someone explain this?

[hem]

one way to look at this problem is (-however, I feel there must be better approach) :...
product of even intergers upto 100
h(100) = xxxxxxxxxx0000 (because it has 100, 10, 50*2, etc..)

So h(100)+1 will be like xxxxxxxx00001

if you try any of the values e.g. 400001, 500001, ...
they are prime numbers or they have prime factor > 40.

hence the answer.

[chix475ntu]

h(100)+1 = 2 * 4 * ..............* 100 +1 = 2^50 * 1 * 2 * .....* 50 + 1

clearly all the prime factors below 50 are factors of h(100), so none of them will be a factor of h(100)+1, so the least will be greater than 50 ==> greater than 40.

E